3.12 Vector applications to kinematicsIB Maths: Applications and Interpretation HL: Revision notes

A body moving with constant velocity $\mathbf{v}$ from initial position $\mathbf{r}_0$ has position vector $\mathbf{r}=\mathbf{r}_0+t\mathbf{v}$ at time $t$. This works in two dimensions and in three. The speed is the magnitude $|\mathbf{v}|=\sqrt{v_1^2+v_2^2+v_3^2}$, and the direction of motion is the direction of $\mathbf{v}$. The path is the straight line through $\mathbf{r}_0$ with direction $\mathbf{v}$. Example: a drone starts at $(2,5)$ with $\mathbf{v}=(3,-4)$ m s$^{-1}$. After 4 s it is at $(2+12,\,5-16)=(14,-11)$, and its speed is $\sqrt{9+16}=5$ m s$^{-1}$. To find when it crosses the $x$-axis, set $5-4t=0$, so $t=1.25$ s.

The position of $B$ relative to $A$ is the vector $\overrightarrow{AB}=\mathbf{r}_B-\mathbf{r}_A$ (end minus start). The distance between them is $|\overrightarrow{AB}|$. If $A$ and $B$ both move at constant velocity, then $\overrightarrow{AB}$ also changes at a constant rate, the relative velocity $\mathbf{v}_B-\mathbf{v}_A$. Example: $\mathbf{r}_A=(1,2)+t(4,1)$ and $\mathbf{r}_B=(9,-1)+t(-2,4)$ (km, hours). Then $\overrightarrow{AB}=(8,-3)+t(-6,3)$. The boats are 5 km apart when $(8-6t)^2+(3t-3)^2=25$, which a GDC solves as $t=0.533$ or $t=2$.

Two paths intersect if the lines cross, so solve $\mathbf{r}_{0}+\lambda\mathbf{v}=\mathbf{r}_{0}'+\mu\mathbf{v}'$ with different parameters $\lambda,\mu$. Use two equations to find $\lambda$ and $\mu$, then check the third in three dimensions. If the check fails, the lines are skew and do not meet. Two objects collide only if they are at the same point at the same time. Use the same $t$ in both position vectors and solve $\mathbf{r}_1(t)=\mathbf{r}_2(t)$. Example: $\mathbf{r}_1=(2,-1,4)+t(1,2,-1)$ and $\mathbf{r}_2=(-1,9,-1)+t(3,-2,1)$. The lines meet at $(5,5,1)$, with $\lambda=3$ and $\mu=2$. Since $D_1$ gets there at $t=3$ and $D_2$ at $t=2$, there is no collision.

For objects with relative position $\overrightarrow{AB}=\mathbf{p}+t\mathbf{q}$, the distance is $d(t)=|\mathbf{p}+t\mathbf{q}|$. To find the closest distance and when it happens, use your GDC to find the minimum of $d(t)$ (or of $d^2$, which has the same minimum) for $t\geq0$. Alternatively, at closest approach the relative position is perpendicular to the relative velocity, so $(\mathbf{p}+t\mathbf{q})\cdot\mathbf{q}=0$. Example (boats): $d^2=45t^2-114t+73$. The minimum is at $t=\frac{19}{15}=1.27$ h, where $d^2=0.8$, so the closest distance is $0.894$ km. Always check $t\geq0$ and give the answer in context, with units.

When the velocity depends on time, e.g. $\mathbf{v}=(7,\,6-4t)$, the motion is no longer a straight line at constant speed. From AHL 5.13: $\mathbf{a}=\frac{d\mathbf{v}}{dt}$ and $\mathbf{r}=\int\mathbf{v}\,dt$, with the constants fixed by the starting position. Speed is $|\mathbf{v}|=\sqrt{v_x^2+v_y^2}$. Example: $\mathbf{a}=(0,-4)$ and, starting at the origin, $\mathbf{r}=(7t,\,6t-2t^2)$. The ball lands when $y=0$: $t=3$, at $x=21$. It is highest when $v_y=0$, so $t=1.5$ and $y=4.5$. Projectile motion is the special case of constant acceleration (here $(0,-4)$): constant horizontal velocity, and a vertical velocity that changes linearly with $t$.

Circular motion is a special case of variable velocity. A point moving anticlockwise on a circle of radius $R$ centred at the origin, with angular speed $\omega$, has $\mathbf{r}=(R\cos\omega t,\ R\sin\omega t)$ and $\mathbf{v}=(-R\omega\sin\omega t,\ R\omega\cos\omega t)$. The speed is constant, $R\omega$, but the velocity keeps changing direction. A time shift describes the same motion starting $a$ seconds later: if $\mathbf{r}(t)$ is the original motion, the delayed motion is $\mathbf{r}(t-a)$ for $t\geq a$. For circular motion $\cos\omega(t-a)$ is a horizontal shift of the cosine graph, i.e. a phase shift (AHL 1.13). Example: a second ball kicked 2 s later has $y=6(t-2)-2(t-2)^2$, equal in height to the first ball when $t=2.5$.